Swim Club Results

Jun 2015
1
0
uk
These are the real results from my kids swim club this morning. Note Lane 2 (of 6) won in 25 of the 45 events.
Am I right that the chances of this are 1 in (6^45)*25/45.

I'm confident about the 6^45 bit being the chance of lane 2 winning each time but is correct to then multiply this by 25/45 to account for winning only 25 out of 45 times?

Is there a better way of analysing them?
 

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mathman

Forum Staff
May 2007
6,933
775
The distribution is binomial. \(\displaystyle \binom{n}{k}p^k(1-p)^{n-k}\).

For your case, n=45, k= 25, p=1/6 (assuming all lanes equally likely).
 
Feb 2010
739
162
These are the real results from my kids swim club this morning. Note Lane 2 (of 6) won in 25 of the 45 events.
Am I right that the chances of this are 1 in (6^45)*25/45.
Assuming that all lanes have an equal chance of winning, then no.

This is a binomial probability problem. The probability of lane 2 winning exactly 25 of 45 races is

\(\displaystyle \binom{45}{25} (1/6)^{25} (5/6)^{20} \approx 3 \times 10^{-11}\)

This is a very small probability. Perhaps it would be more informative to ask "what is the probability of lane 2 winning 25 or more races out of 45".